Al + HF = H2 + AlF3

Al aluminum + HF hydrogen fluoride ⟶ H_2 hydrogen + AlF_3 aluminum fluoride

Balance the chemical equation algebraically: Al + HF ⟶ H_2 + AlF_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 Al + c_2 HF ⟶ c_3 H_2 + c_4 AlF_3 Set the number of atoms in the reactants equal to the number of atoms in the products for Al, F and H: Al: | c_1 = c_4 F: | c_2 = 3 c_4 H: | c_2 = 2 c_3 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 3 c_3 = 3/2 c_4 = 1 Multiply by the least common denominator, 2, to eliminate fractional coefficients: c_1 = 2 c_2 = 6 c_3 = 3 c_4 = 2 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | 2 Al + 6 HF ⟶ 3 H_2 + 2 AlF_3

+ ⟶ +

aluminum + hydrogen fluoride ⟶ hydrogen + aluminum fluoride

| aluminum | hydrogen fluoride | hydrogen | aluminum fluoride molecular enthalpy | 0 kJ/mol | -273.3 kJ/mol | 0 kJ/mol | -1510 kJ/mol total enthalpy | 0 kJ/mol | -1640 kJ/mol | 0 kJ/mol | -3021 kJ/mol | H_initial = -1640 kJ/mol | | H_final = -3021 kJ/mol | ΔH_rxn^0 | -3021 kJ/mol - -1640 kJ/mol = -1381 kJ/mol (exothermic) | | |

Construct the equilibrium constant, K, expression for: Al + HF ⟶ H_2 + AlF_3 Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the activity expression for each chemical species. • Use the activity expressions to build the equilibrium constant expression. Write the balanced chemical equation: 2 Al + 6 HF ⟶ 3 H_2 + 2 AlF_3 Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i Al | 2 | -2 HF | 6 | -6 H_2 | 3 | 3 AlF_3 | 2 | 2 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression Al | 2 | -2 | ([Al])^(-2) HF | 6 | -6 | ([HF])^(-6) H_2 | 3 | 3 | ([H2])^3 AlF_3 | 2 | 2 | ([AlF3])^2 The equilibrium constant symbol in the concentration basis is: K_c Mulitply the activity expressions to arrive at the K_c expression: Answer: | | K_c = ([Al])^(-2) ([HF])^(-6) ([H2])^3 ([AlF3])^2 = (([H2])^3 ([AlF3])^2)/(([Al])^2 ([HF])^6)

Construct the rate of reaction expression for: Al + HF ⟶ H_2 + AlF_3 Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the rate term for each chemical species. • Write the rate of reaction expression. Write the balanced chemical equation: 2 Al + 6 HF ⟶ 3 H_2 + 2 AlF_3 Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i Al | 2 | -2 HF | 6 | -6 H_2 | 3 | 3 AlF_3 | 2 | 2 The rate term for each chemical species, B_i, is 1/ν_i(Δ[B_i])/(Δt) where [B_i] is the amount concentration and t is time: chemical species | c_i | ν_i | rate term Al | 2 | -2 | -1/2 (Δ[Al])/(Δt) HF | 6 | -6 | -1/6 (Δ[HF])/(Δt) H_2 | 3 | 3 | 1/3 (Δ[H2])/(Δt) AlF_3 | 2 | 2 | 1/2 (Δ[AlF3])/(Δt) (for infinitesimal rate of change, replace Δ with d) Set the rate terms equal to each other to arrive at the rate expression: Answer: | | rate = -1/2 (Δ[Al])/(Δt) = -1/6 (Δ[HF])/(Δt) = 1/3 (Δ[H2])/(Δt) = 1/2 (Δ[AlF3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| aluminum | hydrogen fluoride | hydrogen | aluminum fluoride formula | Al | HF | H_2 | AlF_3 Hill formula | Al | FH | H_2 | AlF_3 name | aluminum | hydrogen fluoride | hydrogen | aluminum fluoride IUPAC name | aluminum | hydrogen fluoride | molecular hydrogen | trifluoroalumane

| aluminum | hydrogen fluoride | hydrogen | aluminum fluoride molar mass | 26.9815385 g/mol | 20.006 g/mol | 2.016 g/mol | 83.976748 g/mol phase | solid (at STP) | gas (at STP) | gas (at STP) | solid (at STP) melting point | 660.4 °C | -83.36 °C | -259.2 °C | 1040 °C boiling point | 2460 °C | 19.5 °C | -252.8 °C | density | 2.7 g/cm^3 | 8.18×10^-4 g/cm^3 (at 25 °C) | 8.99×10^-5 g/cm^3 (at 0 °C) | 3.1 g/cm^3 solubility in water | insoluble | miscible | | surface tension | 0.817 N/m | | | dynamic viscosity | 1.5×10^-4 Pa s (at 760 °C) | 1.2571×10^-5 Pa s (at 20 °C) | 8.9×10^-6 Pa s (at 25 °C) | odor | odorless | | odorless |